
Matrices proof
If X = $\displaystyle P^{1}AP and A^{3} = I, prove that X^{3} =I$
I did this
$\displaystyle X^{3} = (P^{1}AP)(P^{1}AP)(P^{1}AP) = (P^{3}A^{3}P^{3}) = (P^{3+3}A^{3}) = A^{3} = I$
I used the law of indices, is it right to apply this for matrices?
Please help me, I am not sure if I did it right here!
Thank you!

Re: Matrices proof
Hey Tutu.
Hint: Group the P*P^(1) = I (I is identity matrix) together to get a lot of cancellation and then use A^3 = I to get your final result.

Re: Matrices proof
Thank you!
I thought I could not change the order within the multiplication, but so is it that I can do it in thie case? Why?
Is it
X^3 = (P^(1)PA)(P^(1)PA)(P^(1)PA)
= P^(3) (PA)^3
= I
Wait, I did this using laws of indices again, can you please explain how to do it, I'm afraid I do not understand..
Thank you ever so much!

Re: Matrices proof
your proof is incorrect.
$\displaystyle X^3 = (P^{1}AP)^3 = (P^{1}AP)(P^{1}AP)(P^{1}AP)$
$\displaystyle = P^{1}A(PP^{1})A(PP^{1})AP = P^{1}AIAIAP = P^{1}AAAP = P^{1}A^3P$
NOT $\displaystyle P^{3}A^3P^3$.
and since $\displaystyle A^3 = I$,
$\displaystyle P^{1}A^3P = P^{1}IP = P^{1}P = I$.
no "switching around of order" is required.

Re: Matrices proof
Thank you!
Does that mean that in matrices, PP^(1) = P? I thought it equalled to I

Re: Matrices proof
Sorry, ignore the previous post. I get it now!